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I am struggeling with an easy deduction that I cannot see for some reason today myself, hope to get some help on this: Suppose $T$ is a compact self-adjoint operator on a Hilbert space $\mathcal{H}$. Then I understand that $(Tx,x)$ is always real, how can I deduce from this that for $\lambda \in \mathcal{C}$

\begin{equation} |((T-\lambda)x,x)| \leq |\text{Im}(\lambda)||x|^2 \end{equation}

Many thanks !!

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Probably that inequality should go in the opposite direction. The good answer below addresses the correct direction of inequality. –  paul garrett Mar 5 at 19:19

1 Answer 1

up vote 5 down vote accepted

\begin{align} ((T-\lambda)x,x) &= (Tx,x)-(\lambda x, x) \\ &= (Tx,x)- \bar \lambda |x|^2\\ &= (Tx,x)-\text{Re}(\lambda)|x|^2+ i\text{ Im}(\lambda)|x|^2 \end{align} The imaginary part of $((T-\lambda)x,x)$ is $\text{Im}(\lambda) |x|^2$. Hence, $|((T-\lambda)x,x)|$ must be at least $|\text{Im}(\lambda)| |x|^2$.

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